How to calulate the Probability(Rain/Falling Leaves) and P(DR) and P(FS)

Question

How to calulate the $P(R\mid FL)$ and $P(DR)$ and $P(FS)$ Please refer the jpeg. Click here for Network Extension (added extra node) to the network in the following stackexchange page: Getting started on a Bayesian network word problem, solving conditional probabilities?

Answer 1

$\def\suml{{\sum}} \def\P{\mathop{\mathbb P}} \def\R{\mathrm R} \def\DR{\mathrm {D\!R}} \def\A{\mathrm A} \def\TS{\mathrm {T\!S}} \def\FL{\mathrm {F\!L}} \def\FS{\mathrm {F\!S}}$Using the convention that $X^\top$ is the event $X$, and $X^{\bot}$ is its complement, and letting $\{\bot,\top\}$ be the implicit interval for summation.

For the simple case of $\P(\R)$, observing that we are given $\P(\TS), \P(\TS^\bot), \P(\R\mid \TS), \P(\R\mid \TS^\bot)$ should make it clear that using the Law of Total Probability is the way to go.   The relevant part of the DAG is simply: $\TS\to \R$.

$$\P(\R) ~{=\suml_{\beta}\P(\TS^\beta, \R) \\= \suml_{\beta} \P(\TS^\beta)\P(\R\mid \TS^\beta)\\ = \P(\TS)\P(\R\mid\TS)+\P(\TS^\bot)\P(\R\mid \TS^\bot)}$$

For the more complicated case of $\P(\R,\FL)$, the nodes of interest are $\R,\FL$ and their parents, $\A,\TS$.   The DAG and the provided probability table indicate how these interrelate. $$\bbox[1ex,border:1pt solid salmon]{\lower{2ex}\R\swarrow \raise{2ex}{\TS}\searrow\lower{2ex} {\FL} \swarrow\raise{2ex} \A}$$

$$\P(\R,\FL) ~{= \suml_{\alpha}\suml_{\beta} \P(\A^\alpha,\TS^\beta, \R,\FL) \\ = \suml_{\alpha}\suml_{\beta}\P(\TS^\beta)\P(\A^\alpha,\R,\FL\mid\TS^\beta) \\= \suml_{\alpha}\suml_{\beta}\P(\TS^\beta)\P(\R\mid \TS^\beta)\P(\FL,A^\alpha\mid\TS^\beta)\\ =\suml_{\alpha}\suml_{\beta}\underline{\phantom{\P(\A^\alpha)\P(\TS^\beta)\P(\R\mid \TS^\beta)\P(\FL\mid \TS^\beta,A^\alpha)}}}$$

You can complete.

The procedure continues likewise for the $\P(\FS)$, and indeed finding the above is a key point.

$$\bbox[1ex,border:1pt solid salmon]{\lower{2ex}\R{\swarrow \raise{2ex}{\TS}\searrow}\hspace{-8ex}{\lower{4ex}\searrow\lower{6ex}{\FS}\lower{4ex}\swarrow}\lower{2ex} {\FL} \swarrow\raise{2ex} \A}$$

$$\P(\FS) ~{ = \suml_{\gamma}\suml_{\delta} \P(\R^\delta,\FL^\gamma)\P(\FS\mid \R^\delta, \FL^\gamma) \\= \suml_{\alpha}\suml_{\beta}\suml_{\gamma}\suml_{\delta} \underline{\phantom{\P(\A^\alpha)\P(\TS^\beta)\P(\FL^\gamma\mid\TS^\beta,\A^\alpha)\P(\R^\delta\mid\TS^\beta)}} \P(\FS\mid \R^\delta, \FL^\gamma)}$$

Now for $\P(\DR)$ the relevant portion of the DAG is just $\TS\to\R\to\DR$.